Question

$2$ moles of an ideal monatomic gas is carried from a state $(P_0, V_0)$ to a state $(2P_0, 2V_0)$ along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

• A. $3P_{0} V_{0}$
• B. $\frac{9}{2}P_{0} V_{0}$
• C. $6P_{0} V_{0}$
• D. $\frac{3}{2}P_{0} V_{0}$

Answer 1

The Correct Option is C

To determine the amount of heat absorbed by a gas as it moves from a state (P_0, V_0) to a state (2P_0, 2V_0) along a straight line in the P-V diagram, we need to understand the process being described and use the first law of thermodynamics. Let's proceed step by step:

1. First, identify that the process goes from (P_0, V_0) to (2P_0, 2V_0). Since it is along a straight line in a P-V diagram, the states are linearly connected, implying a direct relationship between pressure and volume changes.
2. The equation of the straight line can be derived from the endpoints: P = mV + c. Here, m is the slope. By substituting the points:
   • At (P_0, V_0), P_0 = mV_0 + c
   • At (2P_0, 2V_0), 2P_0 = m(2V_0) + c
   Solving these gives m = P_0/V_0 and c = 0.
3. The process is linear from (P_0, V_0) to (2P_0, 2V_0), which means the work done (W) during the process can be calculated as:
   W = \int_{V_0}^{2V_0} P \, dV
   Substitute P = \frac{P_0}{V_0}V:
   W = \int_{V_0}^{2V_0} \frac{P_0}{V_0}V \, dV = \frac{P_0}{V_0} \left[ \frac{V^2}{2} \right]_{V_0}^{2V_0}
   Calculate the integral:
   = \frac{P_0}{V_0} \left( \frac{(2V_0)^2}{2} - \frac{(V_0)^2}{2} \right)
   = \frac{P_0}{V_0} \left( \frac{4V_0^2}{2} - \frac{V_0^2}{2} \right)
   = \frac{P_0}{V_0} \left( \frac{3V_0^2}{2} \right) = \frac{3}{2}P_0 V_0
4. Use the first law of thermodynamics, Q = \Delta U + W, where Q is the heat absorbed, \Delta U is the change in internal energy, and W is work done.
5. For an ideal monatomic gas, the change in internal energy \Delta U is given by:
   \Delta U = \frac{3}{2} n R \Delta T
   As changes are happening in state variables from (2P_0, 2V_0), the temperature ratio (T/T_0)=2 and pressures and volumes both double, hence, use the ideal gas law PV = nRT for calculation.
6. Calculate \Delta U:
   \Delta U = \frac{3}{2}\cdot 2R (2T_0 - T_0) = 3R \cdot T_0
   Now using PV = nRT, we have:
   3R T_0 = 3 P_0 V_0
7. Finally, calculate the heat absorbed:
   Q = \Delta U + W = 3 P_0 V_0 + \frac{3}{2}P_0 V_0 = 6P_0 V_0

Thus, the amount of heat absorbed by the gas during the process is 6P_{0} V_{0}.